Problem 40 - Advanced Engineering Mathematics

Question

In Problems, find a vector function \(\mathbf{r}\) that satisfies the indicated conditions. $$ \begin{aligned} &\mathbf{r}^{\prime \prime}(t)=\sec ^{2} t \mathbf{i}+\cos t \mathbf{j}-\sin t \mathbf{k} ; \\ &\mathbf{r}^{\prime}(0)=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{r}(0)=-\mathbf{j}+5 \mathbf{k} \end{aligned} $$

Step-by-Step Solution

Answer

The vector function is \( \mathbf{r}(t) = (-\ln |\cos t| + t - 1)\mathbf{i} + (-\cos t + t - 1)\mathbf{j} + (\sin t + 4)\mathbf{k} \).

Step 1: Integrate the Second Derivative

To find the vector function \( \mathbf{r}(t) \), we begin by integrating the second derivative \( \mathbf{r}^{\prime\prime}(t) = \sec^2 t \mathbf{i} + \cos t \mathbf{j} - \sin t \mathbf{k} \) to find \( \mathbf{r}^{\prime}(t) \).\( \int \sec^2 t \, dt = \tan t \), \( \int \cos t \, dt = \sin t \), and \( \int -\sin t \, dt = \cos t \). Therefore, \( \mathbf{r}^{\prime}(t) = \tan t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k} + \mathbf{C} \), where \( \mathbf{C} \) is a constant vector.

Step 2: Apply Initial Condition for First Derivative

Use the condition \( \mathbf{r}^{\prime}(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \) to find \( \mathbf{C} \). Substitute \( t = 0 \) into \( \mathbf{r}^{\prime}(t) = \tan t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k} + \mathbf{C} \). This gives us \( 0 = \tan 0 \mathbf{i} + \sin 0 \mathbf{j} + \cos 0 \mathbf{k} = \mathbf{0} + \mathbf{k} \), leading to \( \mathbf{i} + \mathbf{j} + \mathbf{k} = \mathbf{k} + \mathbf{C} \). Solve for \( \mathbf{C} \): \( \mathbf{C} = \mathbf{i} + \mathbf{j} \). Substitute back into \( \mathbf{r}^{\prime}(t) = \tan t \mathbf{i} + \sin t \mathbf{j} + \cos t \mathbf{k} + \mathbf{i} + \mathbf{j} \).

Step 3: Simplify First Derivative

Simplify \( \mathbf{r}^{\prime}(t) \): \( \mathbf{r}^{\prime}(t) = (\tan t + 1)\mathbf{i} + (\sin t + 1)\mathbf{j} + \cos t \mathbf{k} \).

Step 4: Integrate the First Derivative

Integrate \( \mathbf{r}^{\prime}(t) = (\tan t + 1)\mathbf{i} + (\sin t + 1)\mathbf{j} + \cos t \mathbf{k} \) to find \( \mathbf{r}(t) \). \( \int (\tan t + 1) \, dt = \int \tan t \, dt + \int 1 \, dt = -\ln |\cos t| + t \). \( \int (\sin t + 1) \, dt = -\cos t + t \), \( \int \cos t \, dt = \sin t \). Thus, \( \mathbf{r}(t) = (-\ln |\cos t| + t) \mathbf{i} + (-\cos t + t) \mathbf{j} + \sin t \mathbf{k} + \mathbf{D} \).

Step 5: Apply Initial Condition for Function

Use \( \mathbf{r}(0) = -\mathbf{j} + 5\mathbf{k} \) to find \( \mathbf{D} \). Substitute \( t = 0 \) into \( \mathbf{r}(t) = (-\ln |\cos t| + t)\mathbf{i} + (-\cos t + t)\mathbf{j} + \sin t \mathbf{k} + \mathbf{D} \). This gives \( 0 + \mathbf{i} - \mathbf{j} + \mathbf{k} + \mathbf{D} = -\mathbf{j} + 5 \mathbf{k} \). Therefore, \( \mathbf{D} = -\mathbf{i} + 4 \mathbf{k} \).

Step 6: Construct Final Solution

Substitute \( \mathbf{D} = -\mathbf{i} + 4 \mathbf{k} \) into \( \mathbf{r}(t) \). The solution is: \( \mathbf{r}(t) = (-\ln |\cos t| + t - 1)\mathbf{i} + (-\cos t + t - 1)\mathbf{j} + (\sin t + 4)\mathbf{k} \).

Key Concepts

Vector Function

A vector function is a function that takes one or more variables and produces a vector. In this exercise, we deal with a vector function \( \mathbf{r}(t) \) that maps a real number \( t \) to a vector in three-dimensional space. A vector is often represented using standard unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which point in the direction of the x-axis, y-axis, and z-axis, respectively.
A vector function can describe the path of a moving object or any phenomenon that varies in space. Its components are functions of one or more variables. In our case, \( \mathbf{r}(t) \) is a function of \( t \), which could represent time or any other continuous parameter. This makes vector functions very useful in physics for describing motion and other multidimensional phenomena.
Generally, a vector function can have components that are scalar functions themselves. This means the function \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \) is composed of three separate functions \( f(t) \), \( g(t) \), and \( h(t) \), each applied to a different coordinate axis.

Integration

Integration is a fundamental concept in calculus that is used to find antiderivatives or the area under curves. With vector functions, integration can be used to find a function from its derivative. Since a vector function is composed of several component functions, each component is integrated separately.
In this exercise, we start with the second derivative of the vector function \( \mathbf{r}^{\prime\prime}(t) \), and our goal is to find \( \mathbf{r}(t) \). The first step is to integrate \( \mathbf{r}^{\prime\prime}(t) \) to get \( \mathbf{r}^{\prime}(t) \). With integration, we find the antiderivative of each component function:

• The integration of \( \sec^2 t \) gives \( \tan t \).
• The integration of \( \cos t \) gives \( \sin t \).
• The integration of \( -\sin t \) gives \( \cos t \).

Remember that integration yields a constant term \( \mathbf{C} \), representing the unknown constant vector, which is critical in determining the particular solution.

Initial Conditions

Initial conditions are used to find specific solutions to differential equations. Generally, when integrating a derivative, the result includes one or more constants that need to be determined. Initial conditions provide specific values for the function or its derivatives at certain points, allowing us to solve these unknowns.
For the vector function \( \mathbf{r}(t) \), initial conditions were provided for both \( \mathbf{r}^{\prime}(0) \) and \( \mathbf{r}(0) \). These conditions ensure that the final function \( \mathbf{r}(t) \) fits the specific requirements of the problem. For instance:

• The condition \( \mathbf{r}^{\prime}(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \) helps to identify the constant vector \( \mathbf{C} \) after integrating \( \mathbf{r}^{\prime\prime}(t) \).
• Similarly, \( \mathbf{r}(0) = -\mathbf{j} + 5\mathbf{k} \) is used to pinpoint the constant vector \( \mathbf{D} \) after integrating \( \mathbf{r}^{\prime}(t) \).

These conditions are crucial because without them, we would only achieve a general solution rather than the specific solution needed for this problem.

Indefinite Integrals

An indefinite integral, also known as an antiderivative, is the reverse process of differentiation. It represents a family of functions rather than a single function. This is due to the presence of an arbitrary constant or integration constant in the result.
In our vector function context, calculating indefinite integrals allows us to move from \( \mathbf{r}^{\prime\prime}(t) \) to \( \mathbf{r}^{\prime}(t) \), and then from \( \mathbf{r}^{\prime}(t) \) to \( \mathbf{r}(t) \). For example:

• The indefinite integral of \( \sec^2 t \) is \( \tan t + C \), where \( C \) is a constant.
• The indefinite integral of \( \cos t \) is \( \sin t + C \).
• The indefinite integral of \( -\sin t \) is \( \cos t + C \).

These indefinite integrals provide us with the general form of the antiderivatives. To find the particular solution, we apply the initial conditions that allow us to determine the exact constants, completing the transition from a general solution to the specific vector function required by the problem.